Linear Algebra Examples

$- 4 x + y = - 8$ , $- 3 x + 5 y = 3$

Write the system of equations in matrix form.

$[\begin{array}{c} - 4 & 1 & - 8 \\ - 3 & 5 & 3 \end{array}]$

Row reduce.

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Perform the row operation $R_{1} = - \frac{1}{4} R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = - \frac{1}{4} R_{1}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} - \frac{1}{4} R_{1} & - \frac{1}{4} R_{1} & - \frac{1}{4} R_{1} \\ - 3 & 5 & 3 \end{array}]$

$R_{1} = - \frac{1}{4} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{1}{4} R_{1}$ .

$[\begin{array}{c} (- \frac{1}{4}) \cdot (- 4) & (- \frac{1}{4}) \cdot (1) & (- \frac{1}{4}) \cdot (- 8) \\ - 3 & 5 & 3 \end{array}]$

$R_{1} = - \frac{1}{4} R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ - 3 & 5 & 3 \end{array}]$

Perform the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} & 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ (3) \cdot (1) - 3 & (3) \cdot (- \frac{1}{4}) + 5 & (3) \cdot (2) + 3 \end{array}]$

$R_{2} = 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & \frac{17}{4} & 9 \end{array}]$

Perform the row operation $R_{2} = \frac{4}{17} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{4}{17} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ \frac{4}{17} R_{2} & \frac{4}{17} R_{2} & \frac{4}{17} R_{2} \end{array}]$

$R_{2} = \frac{4}{17} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{4}{17} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ (\frac{4}{17}) \cdot (0) & (\frac{4}{17}) \cdot (\frac{17}{4}) & (\frac{4}{17}) \cdot (9) \end{array}]$

$R_{2} = \frac{4}{17} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - \frac{1}{4} & 2 \\ 0 & 1 & \frac{36}{17} \end{array}]$

Perform the row operation $R_{1} = \frac{1}{4} R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = \frac{1}{4} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{1}{4} R_{2} + R_{1} & \frac{1}{4} R_{2} + R_{1} & \frac{1}{4} R_{2} + R_{1} \\ 0 & 1 & \frac{36}{17} \end{array}]$

$R_{1} = \frac{1}{4} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{4} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{1}{4}) \cdot (0) + 1 & (\frac{1}{4}) \cdot (1) - \frac{1}{4} & (\frac{1}{4}) \cdot (\frac{36}{17}) + 2 \\ 0 & 1 & \frac{36}{17} \end{array}]$

$R_{1} = \frac{1}{4} R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & \frac{43}{17} \\ 0 & 1 & \frac{36}{17} \end{array}]$

Use the result matrix to declare the final solutions to the system of equations.

$x = \frac{43}{17}$

$y = \frac{36}{17}$

The solution is the set of ordered pairs that makes the system true.

$(\frac{43}{17}, \frac{36}{17})$

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